Michael breeds chickens and ducks. Last month, he sold $50$ chickens and $30$ ducks for $\$550$. This month, he sold $44$ chickens and $36$ ducks for $\$532$. How much does a chicken cost, and how much does a duck cost? A chicken costs $\$$
Let $x$ represent the cost of a chicken and let $y$ represent the cost of a duck. Since we have two unknowns, we need two equations to find them. Let's use the given information in order to write two equations containing $x$ and $y$. For instance, we are given that last month, Michael sold $\textit{50}$ chickens and $\textit{30}$ ducks for $\$\textit{550}$. How can we model this sentence algebraically? The total cost of chickens Michael sold last month can be modeled by $50x$, and the total cost of ducks he sold last month can be modeled by $30y$. Since these together add up to $550$, we get the following equation: $50x+30y = 550$ We are also given that this month, Michael sold $\textit{44}$ chickens and $\textit{36}$ ducks for $\$\textit{532}$. This can be expressed as: $44x+36y=532$ Now that we have a system of two equations, we can go ahead and solve it! We can now solve the system of equations by the elimination method. Let's manipulate the equations so one of the variables has the same coefficients but with opposite signs. $ \begin{aligned}{-6}\cdot 50x+({-6})\cdot 30y&={-6}\cdot 550\\\\-300x-180y&=-3300\end{aligned}$ $ \begin{aligned} {5}\cdot44x+{5}\cdot 36y&={5}\cdot532\\\\220x+180y&=2660\end{aligned}$ Now we can eliminate $y$ : − 300 x − 180 y + 220 x + 180 y − 80 x + 0 = − 3300 = 2660 = − 640 \begin{aligned}-300x-180y&=-3300\\\\ {+}\ 220x+180y&=2660\\ \hline\\ -80x+0 &=-640 \end{aligned} When we solve the resulting equation, we obtain that $x =8$. Then, we can substitute this into one of the original equations and solve for $y$ to obtain $y=5$. Recall that $x$ denotes the cost of a chicken and $y$ denotes the cost of a duck. Therefore, a chicken costs $\$\textit{8}$ and a duck costs $\$\textit{5}$.